Question

Calculate the pH of a solution that is 3:1 mixture by volume of 0.25M benzoic acid and 0.50M sodium benzoate respectively. The Ka for benzoic acid is 6.28E-5. If the solution in a) was diluted 20x with neutral water, what would be the resulting pH?

Answer 1

The concentration of benzoic acid in the buffer = 3x 0.25/4

and the concentration of sodium benzoate = 1 x0.5/4

ka = 6.28x10^-5  

thus pKa = 4.2

1)The pH of buffer = pKa + log[conjugate base][acid] Hendersen equation

= 4.2 + log [3x 0.25/4 ] /[1 x0.5/4]

= 4.376

2) As the pH of the buffer depends on the ratio of [conjugate acid]/[acid] , on dilution the ratio does not change , hence the pH remains the same on dilution.

AFter diluting 20x , the pH of buffer = 4.376